[TJOI2018]教科书般的亵渎

问题描述

这题的题面真的妙……望大家结合样例正确理解题意……

首先来个样例解释：

样例1
显然，需要两次亵渎
第一次，a从1,2,3,4,6,7,8,9,10，变为0,0,0,0,1,2,3,4,5，对答案的贡献为
1*1+2*2+3*3+4*4+6*6+7*7+8*8+9*9+10*10=360
第二次，a从0,0,0,0,1,2,3,4,5，变为0,0,0,0,0,0,0,0,0，对答案的贡献为
1*1+2*2+3*3+4*4+5*5=55
所以答案为415

样例2
显然，需要三次亵渎
第一次，a从3,4，变为2,3，对答案的贡献为
3*3*3+4*4*4=91
第二次，a从2,3，变为1,2，对答案的贡献为
2*2*2+3*3*3=35
第二次，a从1,2，变为0,0，对答案的贡献为
1*1*1+2*2*2=9
所以答案为135

解题思路

我们发现，问题实际上是求一系列的\(\sum_{i=a}^bi^k\)。那么问题可以转换为求\(\sum_{i=1}^ni^k\)。

我们不妨记\(\sum_{i= 1}^ni^k\)为\(S_{n,k}\)。现在考虑如何求\(S_{n,k}\)。

我们考虑

\[\begin{aligned} (i+1)^{k+1}-i^{k+1}&={k+1 \choose 1}i^k+{k+1 \choose 2}i^{k-1}+...+{k+1\choose k+1}i^0\\ \sum_{i=1}^n((i+1)^{k+1}-i^{k+1})&=(n+1)^{k+1}-1^{k+1}\\ &={k+1\choose 1}\sum_{i=1}^ni^k+{k+1\choose2}\sum_{i=1}^ni^{k-1}+...+{k+1\choose k+1}\sum_{i=1}^n1\\ &={k+1\choose 1}S_{n,k}+{k+1\choose2}S_{n,k-1}+...+{k+1\choose k+1}S_{n,0}\\ S_{n,k}&=\frac{1}{k+1}((n+1)^{k+1}-1-\sum_{i=0}^{k-1}{k+1\choose k+1-i}S_{n,i}) \end{aligned} \]

这样我们就可以在\(O(k^2)\)求出\(S_{n,k}\)。总时间复杂度\(O(m^4)\)。

参考程序

#include <bits/stdc++.h>
#define LL long long
using namespace std;

const LL Mod = 1000000007;
const LL MaxK = 60;
LL C[ MaxK ][ MaxK ];
LL n, m, A[ MaxK ];
LL Sn[ MaxK ], Power[ MaxK ];

void Exgcd( LL a, LL b, LL &x, LL &y ) {
	if( b == 0 ) {
		x = 1; y = 0; return;
	}
	Exgcd( b, a % b, y, x );
	y -= a / b * x;
	return;
}

LL Inv( LL a, LL b ) {
	LL x, y;
	Exgcd( a, b, x, y );
	if( x < 0 ) x += b;
	return x;
}

void Init() {
	C[ 0 ][ 0 ] = 1;
	for( LL i = 1; i <= MaxK - 1; ++i ) {
		C[ i ][ 0 ] = 1;
		for( LL j = 1; j <= i; ++j )
			C[ i ][ j ] = ( C[ i - 1 ][ j - 1 ] + C[ i - 1 ][ j ] ) % Mod;
	}
	return;
}

void Get_Sn( LL n ) {
	memset( Sn, 0, sizeof( Sn ) );
	Sn[ 1 ] = ( n + 1 ) % Mod * ( n % Mod ) % Mod * Inv( 2, Mod ) % Mod;
	Sn[ 0 ] = n % Mod;
	Power[ 0 ] = 1;
	for( LL i = 1; i <= m + 2; ++i )
		Power[ i ] = ( n + 1 ) % Mod * Power[ i - 1 ] % Mod;
	for( LL i = 2; i <= m + 1; ++i ) {
		LL T = 0;
		T = Power[ i + 1 ] - 1;
		T = ( T % Mod + Mod ) % Mod;
		for( LL j = 0; j <= i - 1; ++j ) {
			T = T - C[ i + 1 ][ i + 1 - j ] * Sn[ j ] % Mod;
			T = ( T % Mod + Mod ) % Mod;
		}
		Sn[ i ] = T * Inv( i + 1, Mod ) % Mod;
	}
	return;
}

LL Ask( LL n, LL k ) {
	Get_Sn( n );
	return Sn[ k ];
}

void Work() {
	scanf( "%lld%lld", &n, &m );
	for( LL i = 1; i <= m; ++i ) scanf( "%lld", &A[ i ] );
	sort( A + 1, A + m + 1 );
	++m;
	A[ 0 ] = 0; A[ m ] = n + 1;
	LL k = m;
	LL Ans = 0;
	for( LL i = 1; i <= k; ++i ) {
		for( LL j = i; j <= m; ++j ) {
			Ans = ( Ans + Ask( A[ j ] - 1, k ) - Ask( A[ j - 1 ], k ) ) % Mod;
			Ans = ( Ans + Mod ) % Mod;
		}
		for( LL j = m; j >= i; --j )
			A[ j ] -= A[ i ];
	}
	printf( "%lld\n", Ans );
	return;
}

int main() {
	Init();
	LL Cases;
	scanf( "%lld", &Cases );
	for( ; Cases; --Cases ) Work();
	return 0;
}